Why does the smoothing capacitor of the relay power supply use a large capacity?

[psychepool]

I have not had much experience with making amps that use relays.
But I have curious about it so looked into the schematics of the various amplifiers.
I found that many amps using relays have capacitors over 1000uF capacity in the relay power supply.

Is there a reason to use such a large capacity capacitor?
Are capacitors with capacitances between 10uF and 220uF causing problems?

[Tony Bones]

Think in terms of energy storage, which, for a capacitor, is E = 1/2 * C * V^2 ("one half C V squared".)

So, if the voltage V is big then you only need a little C to get how ever much energy storage you need. On the other hand, if you want to store the same amount of energy with only a few volts, then you need a really big C.

The power supply for a giant radio transmitter might have only 10 or 15uF even though it's giant. That's because it runs on thousands of volts.

Does that make sense?

[psychepool]

Think in terms of energy storage, which, for a capacitor, is E = 1/2 * C * V^2 ("one half C V squared".)

So, if the voltage V is big then you only need a little C to get how ever much energy storage you need. On the other hand, if you want to store the same amount of energy with only a few volts, then you need a really big C.

The power supply for a giant radio transmitter might have only 10 or 15uF even though it's giant. That's because it runs on thousands of volts.

Does that make sense?

I'm not an electric theory major, but I understand what you're trying to say.

I'd like to run a 12V relay through a rectifier diode and regulator on the 6-0VAC tab of the Power Transformer.
(The amp that I planed to make is likely to use approximately 3 to 4 relays.)
I have seen that 1,000uF~6800uF are used depending on the circuit.
The circuits are simmlar but the values are very different.

What capacity is appropriate?

[Tony Bones]

If you're filtering to feed a regulator, then the smaller amount will be fine.

But you won't get 12 VDC by rectifying 6.3 VAC. Is there a particular circuit that you're thinking of copying?

[martin manning]

This should work. Snipped from Dumble 124 schematic in the files section.

[pdf64]

I think that’s a voltage doubling rectifier arrangement, in case anyone else is confused
Beware of using the amp’s regular 6.3V heater supply to power dc supplies; there’s the issue of 0v referencing (ac or dc side only, not both) and that common power tube failure modes connect the heaters to HT. Which isn’t nice for the rectifier and dc circuitry. As owners of virtually irreparable Marshall 6100 where this has happened can testify.

[martin manning]

Yes, a doubler and that’s necessary if you want 12VDC from 6.3VAC. If the 6.3-0 is the filament supply, then the filament string would not be centered on ground, unless the relay supply ground is isolated. 5V relays might be an option using a FWB and a common ground reference, and the heater string would then be centered. Schotky rectifiers and a LDO regulator would be required. If you do not have a dedicated 6.3VAC winding, I would recommend a small auxiliary transformer as shown in the schematic above.

[Stevem]

Please note that a voltage doubling circuit is a bit misnamed as it produces a output of 2.8 times its input.

[pompeiisneaks]

Please note that a voltage doubling circuit is a bit misnamed as it produces a output of 2.8 times its input.

Yup and that's why the LM7812 is in that above circuit. When I built my #124 dumble, I had something like 16 or 17VDC at the input of the regulator, and get nice smooth 12V needed for the relays on the output.

~Phil

[martin manning]

Please note that a voltage doubling circuit is a bit misnamed as it produces a output of 2.8 times its input.

Not at all! It doubles the peak voltage, where the usual HW, FW, or FWB produces 1x the peak voltage.

[Tony Bones]

Please note that a voltage doubling circuit is a bit misnamed as it produces a output of 2.8 times its input.

Yup and that's why the LM7812 is in that above circuit. When I built my #124 dumble, I had something like 16 or 17VDC at the input of the regulator, and get nice smooth 12V needed for the relays on the output.

~Phil

16 or 17 VDC with ripple is probably just about right for an LM7812.

FWIW, I agree that it makes sense to use a separate transformer to power relays and not try to share a winding with heaters. 6 and 12 VAC transformers are inexpensive and small.

[pompeiisneaks]

Please note that a voltage doubling circuit is a bit misnamed as it produces a output of 2.8 times its input.

Yup and that's why the LM7812 is in that above circuit. When I built my #124 dumble, I had something like 16 or 17VDC at the input of the regulator, and get nice smooth 12V needed for the relays on the output.

~Phil

16 or 17 VDC with ripple is probably just about right for an LM7812.

FWIW, I agree that it makes sense to use a separate transformer to power relays and not try to share a winding with heaters. 6 and 12 VAC transformers are inexpensive and small.

I agree as well, I think mine cost like 10 to 12 dollars. The only thing I didn't carefully check was the height of the one I bought, it barely is too tall for the chassis, so it's like 2mm higher than the chassis top, BUT it flexes the bottom of the chassis a tiny bit when I slide it into the cab, and it worked

~Phil

[R.G.]

A little bit of power sIf upply theory:

A rectifier-capacitor filter setup works by the diode conducting only when the incoming AC's instantaneous value is higher than the cap's voltage. So when power is first turned on, the cap is at 0V, and the diodes turn on for nearly all voltages, so the cap charges up to the peak of the incoming AC voltage. Then as the incoming AC voltage starts down from the peak, the cap holds it's charge voltage and the diodes get reverse biased, so the incoming AC has no more effect and the cap just sits at that voltage.

Well, it sits at that peak voltage unless it has a load. Any load current comes out of the cap and makes its voltage ramp down as energy is sucked out of the cap. So the cap voltage droops - until the next rising half cycle of the AC wave. When the next rising AC voltage gets greater than the diode voltage above the cap's drooped voltage, the diode turns on, charges the cap up to the peak again, then turns off.

This charge-to-peak then ramp-down on the cap is what gives the classical sawtooth ripple voltage on AC caps.

If we want DC out of the cap, we'd like this ripple to be "small". The capacitor equation says that I = C * dv/dt, where I is the current into or out of the cap, C is the capacitance, dv = the change in voltage over some time period, and dt is the length of the time period. In AC rectifier conditions, dt is always a bit smaller than one half cycle of the AC power waveform for full wave rectifiers, or one full cycle for (old, weak, silly) half wave rectifiers. So in 60hz countries, dt can be estimated at 8.666mS.

"dv" is how much the cap runs down between charging peaks. The capacitor equation tells us that too. Rearranging it, we get
dv = I *dt/C
Since dt is fixed at the time for one half cycle of the AC power wave, the change in voltage on the cap is purely determined by the ratio of the load current and the capacitance. So - high load currents make the ripple (dv) voltage bigger, high capacitors make it smaller again.

For B+ voltage supplies of 400 V and more, ripple voltages of 30V, 40V ,etc are "small" and the currents in the amp are in the few-hundred milliampere range, so the capacitor needed is small. For low voltage supplies, like a relay at 12Vdc, you have to keep the ripple down to a volt or two, and even for 100ma currents, the capacitor size needed gets BIG.

A regulator gives you a way to let the DC power ripple, but still turn out smooth DC after the regulator.